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CBSEENMA11012922

Prove the following by using the principle of mathematical induction for all space space straight n element of straight N.

1.3 space plus space 2.3 squared space plus space 3.3 cubed space plus space........ space plus space straight n.3 to the power of straight n space equals space fraction numerator left parenthesis 2 straight n minus 1 right parenthesis space 3 to the power of straight n plus 1 end exponent space plus space 3 over denominator 4 end fraction


Solution

Let P(n): 1.3 space plus space 2.3 squared space plus space 3.3 cubed space plus space.......... space plus space straight n.3 to the power of straight n space equals space fraction numerator left parenthesis 2 straight n minus 1 right parenthesis to the power of 3 straight n plus 1 end exponent plus 3 over denominator 4 end fraction
I.         For n = 1,
         P(1) : 1.3 space equals space fraction numerator left square bracket 2 left parenthesis 1 right parenthesis minus 1 right square bracket 3 to the power of 1 plus 1 end exponent plus 3 over denominator 4 end fraction space rightwards double arrow space 3 space equals space fraction numerator 1 left parenthesis 3 squared right parenthesis plus 3 over denominator 4 end fraction space rightwards double arrow space 3 space equals space fraction numerator 9 plus 3 over denominator 4 end fraction space rightwards double arrow space 3 space equals space 3

∴       P(1) is true.
II.      Suppose the statement is true for n = m, straight m element of space straight N
         P(m): space space 1.3 space plus space 2.3 squared plus 3.3 cubed plus..... space plus space straight m.3 to the power of straight m space equals space fraction numerator left parenthesis 2 straight m minus 1 right parenthesis 3 to the power of straight m plus 1 end exponent plus 3 over denominator 4 end fraction     ...(i)
III.     For n = (m + 1),
         P(m + 1) : 1.3 space plus space 2.3 squared plus 3.3 cubed space plus space......... space plus space left parenthesis straight m plus 1 right parenthesis space 3 to the power of straight m plus 1 end exponent space equals space fraction numerator open square brackets 2 left parenthesis straight m plus 1 right parenthesis space minus 1 close square brackets 3 to the power of straight m plus 1 plus 1 end exponent plus 3 over denominator 4 end fraction
or        1.3 space plus space 2.3 squared space plus space 3.3 cubed space plus space....... plus space straight m.3 to the power of straight m plus left parenthesis straight m plus 1 right parenthesis 3 to the power of straight m plus 1 end exponent space equals space fraction numerator left parenthesis 2 straight m plus 1 right parenthesis.3 to the power of straight m plus 2 end exponent plus 3 over denominator 4 end fraction
      From (i),
         WiredFaculty

∴   WiredFaculty
rightwards double arrow space space space fraction numerator left parenthesis 2 straight m minus 1 right parenthesis space 3 to the power of straight m plus 1 end exponent over denominator 4 end fraction plus space 3 over 4 space plus space left parenthesis straight m plus 1 right parenthesis space 3 to the power of straight m plus 1 end exponent space equals space fraction numerator left parenthesis 2 straight m plus 1 right parenthesis 3 to the power of straight m plus 2 end exponent space plus space 3 over denominator 4 end fraction
rightwards double arrow space space space space space space 3 to the power of straight m plus 1 end exponent open square brackets fraction numerator 2 straight m minus 1 over denominator 4 end fraction plus left parenthesis straight m plus 1 right parenthesis close square brackets space plus space 3 over 4 space equals space fraction numerator left parenthesis 2 straight m plus 1 right parenthesis 3 to the power of straight m plus 2 end exponent plus 3 over denominator 4 end fraction
space space rightwards double arrow space space space space 3 to the power of straight m plus 1 end exponent open square brackets fraction numerator 6 straight m plus 3 over denominator 4 end fraction close square brackets space plus space 3 over 4 space equals space fraction numerator left parenthesis 2 straight m space plus space 1 right parenthesis 3 to the power of straight m plus 2 end exponent space plus space 3 over denominator 4 end fraction space rightwards double arrow space fraction numerator 3 to the power of straight m plus 1 end exponent.3 left parenthesis 2 straight m space plus 1 right parenthesis over denominator 4 end fraction space plus 3 over 4 space equals space fraction numerator left parenthesis 2 straight m plus 1 right parenthesis.3 to the power of straight m plus 2 end exponent space plus 3 over denominator 4 end fraction
rightwards double arrow space space space space space space space fraction numerator left parenthesis 2 straight m plus 1 right parenthesis to the power of 3 straight m plus 2 end exponent plus 3 over denominator 4 end fraction space equals space fraction numerator left parenthesis 2 straight m plus 1 right parenthesis 3 to the power of straight m plus 2 end exponent plus 3 over denominator 4 end fraction
         Which is true

∴        P(m + 1) is true

∴        P(m) is true rightwards double arrow P(m + 1) is true
Hence, by the principle of mathematical induction, statement P(n) is true for all straight n space element of space straight N.


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